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CHAPTER 25 Derivatives of Inverse Trig Functions, Lecture notes of Reasoning

Merrimack CollegeReasoning

25.1 Derivatives of Inverse Sine and Cosine. Applying the inverse rule (25.1) with f (x) = sin(x) yields d dxhsin°1(x)i = 1 cos°sin°1(x)¢.

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CHAPTER 25
Derivatives of Inverse Trig Functions
Our
goal is simple, and the answers will come quickly. We will derive six
new derivative formulas for the six inverse trigonometric functions:
d
dxhsin°1(x)id
dxhtan°1(x)id
dxhsec°1(x)i
d
dxhcos°1(x)id
dxhcot°1(x)id
dxhcsc°1(x)i
These formulas will flow from the inverse rule from Chapter 24 (page 278):
d
dxhf°1(x)i=1
f0°f°1(x)¢.(25.1)
25.1 Derivatives of Inverse Sine and Cosine
Applying the inverse rule (25.1) with f(x)=sin(x)yields
d
dxhsin°1(x)i=1
cos°sin°1(x)¢.(25.2)
We are almost there. We just have to simplify the
cos°sin°1(x)¢in the denominator. To do this recall
sin°1(x)=√the angle °º
2∑µ∑º
2
for which sin(µ)=x!.
Thus
sin°1(x)
It is the angle (between
°º
2
and
º
2
) of a the
triangle on the unit circle whose opposite side is
x
. (Be-
cause
sin
of this angle equals
x
.) Then
cos°sin°1(x)¢
is the
length of the adjacent side. By the Pythagorean theorem
this side length is
p1°x2
. Putting
cos°sin°1(x)¢=p1°x2
into the above Equation (25.2), we get or latest rule:
1
sin°1(x)
x
| {z }
cos°sin°1(x)¢
Rule 20 d
dxhsin°1(x)i=1
p1°x2
pf3
pf4
pf5
pf8

Partial preview of the text

Download CHAPTER 25 Derivatives of Inverse Trig Functions and more Lecture notes Reasoning in PDF only on Docsity!

CHAPTER 25

Derivatives of Inverse Trig Functions

O

ur goal is simple, and the answers will come quickly. We will derive six

new derivative formulas for the six inverse trigonometric functions:

d

dx

h sin

° 1 (x)

i (^) d

dx

h tan

° 1 (x)

i (^) d

dx

h sec

° 1 (x)

i

d

dx

h cos

° 1 (x)

i (^) d

dx

h cot

° 1 (x)

i (^) d

dx

h csc

° 1 (x)

i

These formulas will flow from the inverse rule from Chapter 24 (page 278):

d

dx

h f

° 1 (x)

i

f 0

f °^1 (x)

25.1 Derivatives of Inverse Sine and Cosine

Applying the inverse rule (25.1) with f (x) = sin(x) yields

d

dx

h sin

° 1 (x)

i

cos

sin

° 1 (x)

We are almost there. We just have to simplify the

cos

sin

° 1 (x)

in the denominator. To do this recall

sin°^1 (x) =

the angle ° º 2 ∑ μ ∑ º 2

for which sin( μ ) = x

Thus sin

° 1

(x) It is the angle (between °

º

2 and^

º

2 ) of a the

triangle on the unit circle whose opposite side is x. (Be-

cause sin of this angle equals x.) Then cos

sin

° 1 (x)

is the

length of the adjacent side. By the Pythagorean theorem

this side length is

p

1 ° x^2. Putting cos

sin

° 1 (x)

p 1 ° x^2

into the above Equation (25.2), we get or latest rule:

1

sin°^1 (x)

x

| {z }

cos

sin°^1 (x)

Rule 20

d

dx

h sin

° 1 (x)

i

p 1 ° x^2

Derivatives of Inverse Sine and Cosine 287

We reviewed sin°^1 (x) In Section 6.1 and presented its graph on page 101.

Figure 25.1 repeats the graph, along with the derivative from Rule 20.

x

y

  • 1 1

1

° º 2

º 2 f^ (x)^ =^ sin

° 1 (x)

f

0 (x) =

p 1 ° x^2

f (x) = cos°^1 (x)

Figure 25.1. The function sin°^1 (x) and its derivative. The derivativeis

always positive, reflecting the fact that the tangents to sin

° 1

(x) have positive

slope. The derivative has vertical asymptotes at x = ± 1 , as the tangents to

sin

° 1

(x) become increasingly steep as x approaches ± 1.

Now consider cos°^1 (x). The tangents to its graph (Figure 25.2 below)

have negative slope, and the geometry suggests that its derivative is negative

the derivative of sin

° 1

(x). Indeed this turns out to be exactly the case. This

chapter’s Exercise 1 asks you to prove our next rule:

Rule 21

d

dx

h cos°^1 (x)

i

p 1 ° x^2

x

  • 1 1 ° 1

f (x) = cos°^1 (x)^ º

f

0 (x) =

p 1 ° x^2

Figure 25.2. The function cos°^1 (x) and its derivative.

Derivatives of Inverse Secant and Cosecant 289

25.3 Derivatives of Inverse Secant and Cosecant

We reviewed sec°^1 (x) in Section 6.3. For its derivative, put f (x) = sec(x) into

the inverse rule (25.1), with f °^1 (x) = sec°^1 (x) and f 0 (x) = sec(x) tan(x). We get

d

dx

h sec

° 1 (x)

i

f 0

f °^1 (x)

sec

f °^1 (x)

· tan

f °^1 (x)

sec

sec°^1 (x)

· tan

sec°^1 (x)

Because sec

sec

° 1 (x)

= x, the above becomes

d

dx

h sec

° 1 (x)

i

x · tan

sec°^1 (x)

In Example 6.5 we showed that tan

sec

° 1 (x)

( (^) p

x^2 ° 1 if x is positive

p

x^2 ° 1 if x is negative

With this, Equation 25.4 above becomes

d

dx

h sec

° 1 (x)

i

x

p x^2 ° 1

if x is positive

°x

p x^2 ° 1

if x is negative.

But if x is negative, then °x is positive, and the above consolidates to

Rule 24

d

dx

h sec

° 1 (x)

i

|x|

p x^2 ° 1

This graph of sec°^1 (x) and its derivative is shown in Figure 25.3.

y = f

0 (x) =

|x|

p x^2 ° 1

y

º

º 2 y^ =^ f^ (x)^ =^ sec

° (^1) (x)

° 1 1

Figure 25.4. The graph of sec°^1 (x) and its derivative. The domain of

both functions is (°1, °1] [ [1, 1 ). Note that the derivative has vertical

asymptotes at x = ± 1 , where the tangent line to y = sec°^1 (x) is vertical.

290 Derivatives of Inverse Trig Functions

This chapter’s Exercise 2 asks you to use reasoning similar to the above

to deduce our final rule.

Rule 25

d

dx

h csc°^1 (x)

i

|x|

p x^2 ° 1

Each of our new rules has a chain rule generalization. For example,

Rule 25 generalizes as

d

dx

h csc°^1

g(x)

¢i

Ø

Øg(x)

Ø

Ø

q

(g(x))

g^0 (x) =

°g^0 (x) Ø Øg(x)

Ø

Ø

q

(g(x))

Here is a summary of this Chapter’s new rules, along with their chain rule

generalizations.

d

dx

h sin

° 1 (x)

i

p 1 ° x^2

d

dx

h sin

g(x)

¢i

q

1 ° (g(x))

2

g

0 (x)

d

dx

h cos

° 1 (x)

i

p 1 ° x^2

d

dx

h cos

g(x)

¢i

q

1 ° (g(x))

2

g

0 (x)

d

dx

h tan°^1 (x)

i

1 + x^2

d

dx

h tan°^1

g(x)

¢i

1 + (g(x))

2

g^0 (x)

d

dx

h cot°^1 (x)

i

1 + x^2

d

dx

h cot°^1

g(x)

¢i

1 + (g(x))

2

g^0 (x)

d

dx

h sec°^1 (x)

i

|x|

p x^2 ° 1

d

dx

h sec°^1

g(x)

¢i

g

0 (x) Ø Øg(x)

Ø

Ø

q

(g(x))

d

dx

h csc

° 1 (x)

i

|x|

p x^2 ° 1

d

dx

h csc

g(x)

¢i

°g^0 (x) Ø Øg(x)

Ø

Ø

q

(g(x))^2 ° 1

Example 25.

d

dx

hp cos°^1 (x)

i

d

dx

h° cos

° 1 (x)

2

i

cos

° 1 (x)

2 d dx

h cos

° 1 (x)

i

cos

° 1 (x)

(^2) p°^1

1 ° x^2

p cos°^1 (x)

p 1 ° x^2

Example 25.

d

dx

h e

tan°^1 (x)

i = e

tan°^1 (x) d dx

h tan

° 1 (x)

i = e

tan°^1 (x) 1 1 +x^2

etan

° (^1) (x)

1 +x^2

Example 25.

d

dx

h tan°^1

ex

¢ i

1 + (^) (ex)^2

d

dx

h ex

i

1 + e^2 x^

ex^ =

e

x

1 + e^2 x^

292 Derivatives of Inverse Trig Functions

In addition we have the following general rules for the derivatives of

combinations of functions.

Constant Multiple Rule:

d

dx

c f (x)

= c f 0 (x)

Sum/Dierence Rule:

d

dx

f (x) ± g(x)

= f 0 (x) ± g^0 (x)

Product Rule:

d

dx

f (x)g(x)

= f 0 (x)g(x) + f (x)g^0 (x)

Quotient Rule:

d

dx

f (x)

g(x)

f 0 (x)g(x) ° f (x)g^0 (x)

(g(x))^2

Chain Rule:

d

dx

f

g(x)

= f

0 (g(x)) g

0 (x)

Inverse Rule:

d

dx

f

° 1 (x)

f 0

f °^1 (x)

We used this last rule, the inverse rule, to find the derivatives of ln(x)

and the inverse trig functions. After it has served these purposes it is mostly

retired for the remainder of Calculus I, except for the stray exercise or quiz

or test question.

This looks like a lot of rules to remember, and it is. But through practice

and usage you will reach the point of using them automatically, with hardly

a thought. Be sure to get enough practice!

Exercises for Chapter 25

1. Show that

d

dx

h cos°^1 (x)

i

p 1 ° x^2

2. Show that

d

dx

h csc

° 1 (x)

i

|x|

p x^2 ° 1

3. Show that

d

dx

h cot

° 1 (x)

i

1 + x^2

Find the derivatives of the given functions.

4. sin

° 1 °p 2 x

5. ln

tan°^1 (x)

6. e

x tan

° 1 (x)

7. tan°^1

º x

8. sec°^1

º x

9. ln

sin

° 1 (x)

10. cos°^1

º x

11. sec°^1

x^5

12. e

tan°^1 ( º x)

13. tan

° 1 °^

ln(x)

+ º 14. tan

x sin(x)

15. x sin

° 1 °^

ln(x)

Summary of Derivative Rules 293

Exercise Solutions for Chapter 25

1. Show that

d

dx

h cos

° 1 (x)

i

p 1 ° x^2

By the inverse rule,

d

dx

h cos

° 1 (x)

i

° sin

cos°^1 (x)

cos°^1 (x) º (^) 0

1

Now we simplify the denominator.

From the standard diagram for

cos°^1 (x) we get sin

cos°^1 (x)

OPP HYP

p 1 °x^2 1 =^

p

1 ° x^2. With this, the above

becomes

d

dx

h cos

° 1 (x)

i

p 1 ° x^2

x

3. Show that

d

dx

h cot°^1 (x)

i

1 + x^2

Suggestion: Verify the identity cot°^1 (x) = º 2 ° tan°^1 (x). Then dierentiate

both sides of this.

d

dx

h ln

tan°^1 (x)

¢i

tan°^1 (x)

d

dx

h tan°^1 (x)

i

tan°^1 (x)

1 + x^2

tan°^1 (x)

1 + x^2

d

dx

h tan

º x

¢i

º

1 + ( º x)^2

º

1 + º^2 x^2

d

dx

h ln

sin°^1 (x)

¢ i

sin

° 1 (x)

d

dx

h sin°^1 (x)

i

sin

° 1 (x)

p 1 °x^2

sin

° 1 (x)

p 1 °x^2

d

dx

h sec

x

5 ¢^

i

Ø

Øx^5

Ø

Ø

q ° x^5

5 x

4

5 x

4 Ø Øx^5

Ø

Ø

p x^10 ° 1

Ø

Øx

Ø

Ø

p x^10 ° 1

d

dx

h tan

° 1 °^

ln(x)

i

ln(x)

x

x + x

ln(x)

d

dx

h x sin

° 1 °^

ln(x)

¢i = 1 · sin

° 1 °^

ln(x)

  • x

d

dx

h sin

° 1 °^

ln(x)

¢i

= sin

° 1 °^

ln(x)

  • x

q 1 °

ln(x)

x

= sin

° 1 °^

ln(x)

q 1 °

ln(x)